Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Answer:
Good question to ask in physics, sir maam
The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let 
be the masses of the three fragments.
Let 
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,
So the x- component of the velocity of the m2 fragment after the explosion is,
∴ 
210 Pb ---> -ie + 210 B:
84 8.3
